Determining the result or state of system in a calorimeter

1. The problTfem statement, all variables and given/known data
Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water.

2. Relevant equations
FInding the mass. I think I figured out the final temp?

3. The attempt at a solution
Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat

1. The problTfem statement, all variables and given/known data
Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water.

2. Relevant equations
FInding the mass. I think I figured out the final temp?

3. The attempt at a solution
Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.

I am struggling to find the grams of steam condensed.

m(-540g/C)=Ice melted + Rise in water?

Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.

Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.

Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

Also, this time around, you don't need to worry about the portion of water, which was once steam, dropping to some final temperature. The Final temperature is 100 oC. Yes, there is steam that condenses into water, but it doesn't change temperature from there. So that simplifies things.

Also, this time around, you don't need to worry about the portion of water, which was once steam, dropping to some final temperature. The Final temperature is 100 oC. Yes, there is steam that condenses into water, but it doesn't change temperature from there. So that simplifies things.

When we look at hese final systems in both instances how do we know what to pick? IN terms of ice or steam mass to find?

In the first case the final temperature is 149C or so. SInce it is water it cant go higher. so it is really 100C. How do we know to analyze steam? Does it mean that the stuff in the calorimeter GAINED some energy? From the steam since there was so much?

In the second the negative number means that the system or the water LOST energy to the ice? Melting it?

Its like depending on how much you have one steam or ice buffers the temp? THanks

I feel like conceptually I am still quite not getting it I have a feeling it depends on what that final temperature number means in all three states of water in the system and who is exchanging with what...

WHat I need to do is look in each case are the energies needed to metl ice and condense steam. For instance in the the first example the energies for the ice and steam are 1600 cal and 54000 cal respectively. Also to raise the contents(calorimeter mass + water + ice water mass) with steam takes 250g(100C)(1cal/gC) =26,000cal(this is the energy at equilibrium the ice would have to melt so it would all be water) Since the steam is way larger than the system all of it melts and manages only to condense some of the steam raising the while raising the water . This means that we can find out how much steam is condensed:

m(-540g/C)= 26,600. m=49.

The way I ws doing it before made it difficult to "pick" which one I needed to find the mass for. We can get the final temp but I don't know what it means for each unless I could quantitiatively compare the energies of each component(Contents of total calorimeter ice vs steam.).

"Ice vs steam" tells me which one will be left over. Total calorimeter tells me what final engery we have to find the mass of said state of water left over.

Unless someone can tell me how to do so with the way I did it BEFORE this?

The energy of the steam condensing must be equal to the energy of the ice melting, if the final state is ice and water.

The energy of the steam condensing must be equal to the energy of the ice melting, plus the energy of the water (some of it previously ice) and calorimeter changing temperature, if the final state is all water (and calorimeter), where the final temperature is in between 0 and 100 deg C.

Same as above if the final temperature is 100 deg C, except in this case, not all the steam condenses.

Conservation of Energy applies in any case.

So how do you which one is the final state? Well, there's a couple of ways. You could try it one way and see if the results make sense. If they don't try it a different way. Or, you could calculate the energies of certain, particular changes and compare them, for example comparing the energy of the total steam condensing compared to the energy of the ice melting. That can at least help narrow down the possibilities.

My approach I think is I will compare the energies of steam and ice so like in the case of the first the steam is larger. That allows me to identify that the system is heating up to 100C(stuff in calorimeter) and that since one is larger the ice melts. This means that essentially the final temp is 100.
SInce Qlost=Qgained:

M(540g/C)= basically the rest of the systems compontents.

The one that exhibits no change in temp is the steam water itself and that term will go to zero.

with a change in temp of 100 degrees plus the energy of ice melting it comes to equaling 26600 cals

so mass is what we said before

This works for the other case when the system COOLS the ice wins eats up all the steam and we can the amount of ice melted.

THis makes more sense to me because it depends on which way we are traveling on the graph or trend. It also allows me to remember that steam(ice) and water can exist at 100C(0C for ice).

So identify what happens to ice and steam. Then assess if heating or cooling to a final temp then solve accordingly.